hdu 5191 Building Blocks（前缀和）

Building Blocks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1227    Accepted Submission(s): 266

Problem Description

After enjoying the movie,LeLe went home alone. LeLe decided to build blocks.
LeLe has already built n piles. He wants to move some blocks to make W consecutive piles with exactly the same height H.

LeLe already put all of his blocks in these piles, which means he can not add any blocks into them. Besides, he can move a block from one pile to another or a new one，but not the position betweens two piles already exists.For instance,after one move,"3 2 3" can become "2 2 4" or "3 2 2 1",but not "3 1 1 3".

You are request to calculate the minimum blocks should LeLe move.

 

Input

There are multiple test cases, about 100 cases.

The first line of input contains three integers n,W,H(1≤n,W,H≤50000).n indicate n piles blocks.

For the next line ,there are n integers A1,A2,A3,……,An indicate the height of each piles. (1≤Ai≤50000)

The height of a block is 1.

 

Output

Output the minimum number of blocks should LeLe move.

If there is no solution, output "-1" (without quotes).

 

Sample Input

4 3 21 2 3 54 4 41 2 3 4

 

Sample Output

1-1
Hint
In first case, LeLe move one block from third pile to first pile.
 

 

Source

BestCoder Round #34

题意：有好多堆，可以将一堆中取出一个块放到别的堆，或者放到两端形成新的堆，问最少挪几次能形成连续的宽度为w高为h的一排堆
题目分析：记录两个前缀和，一个是需要取走的块数，一个是总的块数，枚举起点，然后对w区间进行处理即可
 

#include <iostream>#include <cstring>#include <algorithm>#include <cstdio>#define MAX 400007using namespace std;typedef long long LL;LL n,w,h;int a;LL out[MAX];LL sum[MAX];int main ( ){    while ( ~scanf ( "%I64d%I64d%I64d" , &n , &w , &h ) )    {        LL ans = 1234567891234;        memset ( out , 0 ,sizeof ( out ) );        memset ( sum , 0 , sizeof ( sum ) );        for ( int i = w+1 ; i <= w+n; i++ )        {            scanf ( "%d" , &a );            out[i] = out[i-1];            if ( a > h ) out[i] += a-h;            sum[i] = sum[i-1] + a;                       }        for ( int i = w+n+1 ; i <= n+2*w+2 ; i++ )            out[i] = out[i-1],sum[i] = sum[i-1];        if ( sum[n+2*w] < w*h )        {            puts ( "-1" );            continue;        }        for ( int i = 1 ; i <= w+n ; i++ )        {            LL sum1 = sum[i+w-1] - sum[i-1];            LL out1 = out[i+w-1] - out[i-1];            if ( sum1 >= w*h )  ans = min ( ans , out1 );            else ans = min ( ans , w*h - sum1 + out1 );        }        printf ( "%I64d\n" , ans );    }}